STRUCTURE OF Na-23 AND Ne-23
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper "Nuclear structure is governed by the fundamental laws of electromagnetism " (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Structure of Na23 ' Using the first diagram of the structure of Na23 which represents the Fig.8e of my published paper one sees that here the core is not the parallelepiped of O-16 with S = 0 but a greater parallelepiped including five horizontal squares with S = +2, because the nucleons of the fifth horizontal square (n9p9n10p10) in the fifth horzontal plane (HP5) give S = +1/2 +1/2 +1/2 +1/2 = +2 . Here the three extra nucleons like the n11, p11, and n12, form the extra rectangles with the three stable radial bonds of n11p4, p11n6 and n12p8 because at points p4 , n6 , and p8 there exist five pn bonds per nucleon which contribute to the increase of the binding energies. In this case as shown also in Fig 8e of my published paper one observes the two extra axial bonds of n11p11 and p11n12 systems. In other words at p11 we see that there exist the three bonds like p11n6, p11n11 and p11n12 which are responsible for the stability of Na23. Under this condition all extra pn bonds overcome the pp and nn repulsions for the formation of the stable Na-23 These extra nucleons give a spin S = -1/2 +1/2 -1/2 = -1/2. Since the spin of the fifth square is S = +2 on gets the total spin of Na23 as S = +2 -1/2 = +3/2. ' ' ' Stable Na-23 of S = +3/2 Unstable Ne-23 of S = +5/2 p10(+1/2.n10(+1/2) p10(+1/2)n10(+1/2) ' '''n9(+1/2).p9(+1/2) n9(+1/2).p9(+1/2) HP5 n8(-1/2).. p8(-1/2)…..n12(-1/2) n8(-1/2).p8(-(1/2) p7(-1/2).n7(-1/2) n13(-1/2)..p7(-1/2).n7(-1/2) HP4 p6(+1/2).n6(+1/2)…..p11(+1/2) p6(+1/2).n6(+1/2) n5(+1/2).p5(+1/2) n5(+1/2).p5 (+1/2) …..n12(+1/2) HP3 n4(-1/2).p4(-1/2)……n11(-1/2) n4(-1/2).p4(-1/2) p3 (-1/2).n3 (- 1/2) p3(-1/2).n3(-1/2) HP2 p2(+1/2).n2(+1/2) p2(+1/2).n2(+1/2) n1(+1/2). p1(+1/2) n1(+1/2).p1(+1/2)…. n11(+1/2) HP1 '''Why the unstable Ne-23 with S = +5/2 turns to the stable Na-23 with S = +3/2 ' ''' '''Looking carefully at the second diagram of Ne-23 one concludes that the core with S = +2 is the same as that of the structure of Na-23. Since the extra neutrons like n11(+1/2) , n12(+1/2) and the n13 (-1/2) give S = +1/2 +1/2 -1/2 = +1/2 we get the total spin of Ne-23 as S = +2 +1/2 = +5/2 But here the structure is unstable because the three extra neutrons form only single np bonds. Especially the neutron n11 forming the single bond of n11p1 is the weaker bond of the structure of Ne-23 because at point p1 there exist only four pn bonds per nucleon, while at points p5 and p7 there exist five pn bonds per nucleon. Under this condition the neutron n11 turns to p11 as n11 = p11 + electron + antineutrino Therefore the three extra single bonds of the three neutrons n11, n12 and n13 turn to the stable axial and radial bonds of the three nucleons n11, p11, and n12 with S = -1/2 for the formation of the stable structure of the Na-23 with spin S = +3/2. Category:Fundamental physics concepts